山东第六届省赛Circle of Friends(强联通缩点最短路)

来源：互联网发布：sql查看创建表语句编辑：程序博客网时间：2024/05/21 09:50

题意：

朋友圈，如果彼此都认为对方是朋友帮忙就无花费，否则花费为1，如果是一个环也认为彼此是朋友。

思路：

强联通分量缩点，重新造边，跑最短路。

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <queue>#include <vector>using namespace std;const int MAXN =100000 ;const int MAXM =100000 ;using namespace std;const int maxn=100005;struct node ///优先队列， 在距离相同时选最小花费{    int  v,w,dis;    node(int _v=0,int _w=0,int _dis=0):v(_v),w(_w),dis(_dis){}    bool operator <(const node &r)const    {        if(dis!=r.dis)            return dis>r.dis;        else            return w>r.w;    }};vector<node> vec[maxn];///存边struct Edge{    int to,next;}edge[MAXM];int head[MAXN],tot;int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];int Index,top;int scc;int vis[maxn];int dis[maxn];///存最短路径const int inf=0x3f3f3f3f;bool Instack[MAXN];void addedge(int u,int v){    edge[tot].to = v;edge[tot].next = head[u];head[u] = tot++;}void dijk(int st,int n){    for(int i=1;i<=n;i++)    {        vis[i]=0;        dis[i]=inf;    }    priority_queue<node> q;    dis[st]=0;    q.push(node(st,0,0));    node tmp;    while(!q.empty())    {        tmp=q.top();        q.pop();        int u=tmp.v;        if(vis[u])            continue;        vis[u]=1;        for(int i=0;i<vec[u].size();i++)        {            int v=vec[u][i].v;            int cost=vec[u][i].w;            if(!vis[v]&&dis[v]>=dis[u]+cost)            {                dis[v]=dis[u]+cost;                q.push(node(  v,cost,dis[v]));            }        }    }    if(dis[Belong[n]]!=inf)        printf("%d\n",dis[Belong[n]]);    else        printf("-1\n");}void Tarjan(int u){    int v;    Low[u] = DFN[u] = ++Index;    Stack[top++] = u;    Instack[u] = true;    for(int i = head[u];i != -1;i = edge[i].next)    {        v = edge[i].to;        if(!DFN[v])        {            Tarjan(v);            if(Low[u] > Low[v])                  Low[u] = Low[v];        }        else if(Instack[v] && Low[u] > DFN[v])            Low[u] = DFN[v];    }    if(Low[u] == DFN[u])    {        scc++;        do        {            v = Stack[--top];            Belong[v] = scc;            Instack[v] = false;        }        while( v!= u );    }}int in[MAXN],out[MAXN];void solve(int N){    for(int i=1;i<=N;i++)              vec[i].clear();      memset(DFN,0,sizeof(DFN));    memset(Instack,false,sizeof(Instack));    Index = scc = top = 0;    for(int i = 1;i <= N;i++)        if(!DFN[i])           Tarjan(i);    for(int u = 1;u <= N;u++)    {        for(int i = head[u];i != -1;i = edge[i].next)        {            int v = edge[i].to;            if(Belong[u] != Belong[v])            {                vec[Belong[u]].push_back(node(Belong[v],1,0));            }        }    }    /*    for(int i=1;i<=scc;i++)    {        for(int j=0;j<vec[i].size();j++)        {            printf("%d--%d\n ",i,vec[i][j]);        }    }*/    dijk(Belong[1],N);}void init(){     tot = 0;    memset(head,-1,sizeof(head));}int main(){    int n,m;    int v;    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        init();        int u,v;        for(int i = 1;i <= m;i++)        {            scanf("%d%d",&u,&v);            u++,v++;            addedge(u,v);        }        solve(n);    }    return 0;}

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